The graph ln K(eq.) vs 1T relates for a reaction. The reaction must be

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Exothermic

Endothermic

∆H is negligible

High spontaneous at ordinary temperature

answer is A.

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Detailed Solution

ln K2K1 = ∆HR T2 - T1T1 T2 Since K is increased or decreased with temperature, thus ∆H = -ve

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